If A = 580^circ,then which one of the followi | vedclass

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(C) Given $A = 580^\circ$. We know that $\sqrt{1 + \sin A} = \sqrt{(\sin(A/2) + \cos(A/2))^2} = |\sin(A/2) + \cos(A/2)|$. And $\sqrt{1 - \sin A} = \sq

Vedclass · Accepted Answer

$2 \sin \left( \frac{A}{2} \right) = -\sqrt{1 + \sin A} - \sqrt{1 - \sin A}$

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(C) Given $A = 580^\circ$. We know that $\sqrt{1 + \sin A} = \sqrt{(\sin(A/2) + \cos(A/2))^2} = |\sin(A/2) + \cos(A/2)|$. And $\sqrt{1 - \sin A} = \sqrt{(\sin(A/2) - \cos(A/2))^2} = |\sin(A/2) - \cos(A/2)|$. For $A = 580^\circ$,$A/2 = 290^\circ$. In the fourth quadrant $(270^\circ Thus,$\sin(290^\circ) + \cos(290^\circ) > 0$ (since $\cos(290^\circ) > |\sin(290^\circ)|$ is false,let's re-evaluate: $\sin(290^\circ) \approx -0.94$,$\cos(290^\circ) \approx 0.34$,so $\sin(290^\circ) + \cos(290^\circ) Actually,$\sin(290^\circ) - \cos(290^\circ) Therefore,$\sqrt{1 + \sin A} = -(\sin(A/2) + \cos(A/2))$ and $\sqrt{1 - \sin A} = -(\sin(A/2) - \cos(A/2))$. Adding these: $\sqrt{1 + \sin A} + \sqrt{1 - \sin A} = -\sin(A/2) - \cos(A/2) - \sin(A/2) + \cos(A/2) = -2 \sin(A/2)$. Thus,$2 \sin(A/2) = -(\sqrt{1 + \sin A} + \sqrt{1 - \sin A})$.

If $A = 580^\circ$,then which one of the following is true?

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